Continuous Random Variable – PDF Normalization and Moments

00:00 19 marks

A continuous random variable \(X\) has a probability density function (PDF) given by
\[
f(x) = \begin{cases} k\,\sin\left(\frac{\pi x}{4}\right) & \text{for } 0 \le x \le 4, \\ 0 & \text{otherwise} \end{cases}
\]

(a) Show that \(k = \frac{\pi}{8}\). [3]

(b) Find \(E(X)\). [4]

(d) Find the variance of \(X)\). [5]

Let \(Y\) be the random variable defined by \(Y = 2X + 3\).

(e) Find \(E(Y)\). [2]

(f) Find \(\operatorname{Var}(Y)\). [2]

Hint Hint

Hint

For part (a), set up the normalization condition \(\int_0^4 k\,\sin(\frac{\pi x}{4}) dx = 1\) and solve for k. For parts (b) and (d), set up the integrals for \(E(X)\) and \(E(X^2)\) respectively, then use \(\operatorname{Var}(X) = E(X^2) – [E(X)]^2\). Remember for parts (e) and (f) to use linearity of expectation and the variance scaling rule, i.e. \(E(2X+3)=2E(X)+3\) and \(\operatorname{Var}(2X+3)=4\operatorname{Var}(X)\).

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Complete Answer

The solution proceeds by first normalizing the PDF to obtain k = π/8. Then, E(X) is determined by evaluating the integral \(\int_0^4 x (\pi/8)\sin(\pi x/4)dx\) and is found to be 2. The median is found by solving \(\int_0^m (\pi/8)\sin(\pi x/4)dx = 0.5\), which gives m = 2. Evaluating \(E(X^2)\) leads to a variance of approximately 1.25. Finally, for Y = 2X + 3, using linearity of expectation and the variance scaling rule, we have E(Y) = 7 and Var(Y) ≈ 5.01.

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